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Thermodynamics Cv = Cp + R Question

  1. Oct 30, 2004 #1
    Can someone tell me why Cv = Cp + R
     
  2. jcsd
  3. Oct 31, 2004 #2
    Well first of all its Cp - Cv = nR, where is your moles of gas, and its only true for an ideal gas. You find this about about halfway through the first semester of physical chemistry.

    The formal definitions of Cv and Cp are

    [tex] C_v = \frac{\partial U}{\partial T} [/tex] and

    [tex] C_p = \frac{\partial H}{\partial T}[/tex]

    Where U is the internal energy and H is the enthalpy, defined to be H = U + pV. But for an ideal gas, pV = nRT. Substitutiotn this into the defition for Cp we get

    [tex] C_p = \frac{\partial (U + nRT)}{\partial T}[/tex]

    [tex] C_p = \frac{\partial U }{\partial T} + \frac{\partial (nRT)}{\partial T}[/tex]

    [tex] C_p = \frac{\partial U}{\partial T} + nR[/tex]

    [tex] C_p = C_v + nR[/tex]
     
    Last edited: Oct 31, 2004
  4. Oct 31, 2004 #3
    What are on?
     
  5. Oct 31, 2004 #4
    Are you talking about [tex]\partial[/tex] ? Thats the symbol for a partial deriviative. Its like a deriviative but for functions of mroe than one variable. To calculate it, you treat the other variables as constants, except for the one youare differentiating against.
     
  6. Oct 31, 2004 #5
    Can u prove the equation by the consideration of an isobaric process?
     
  7. Oct 31, 2004 #6
    No, path has nothing to do with it. It follows directly from the definition of U, H, Cp and Cv, and our assumption of an ideal gas. If our gas it is not ideal, that equation does not hold.
     
  8. Oct 31, 2004 #7
    But what if we consider an ideal gas undergoing a isochoric process and
    how the first law of thermodynamics applies to it.
    ΔU= Q - W

    Since the work is defined by the pressure * the change in volume
    W = pΔV
    then work is 0

    Hence ΔU = Q - 0 = Q

    but Q = nCvΔT hence ΔU = nCvΔT

    But since the change in internal energy is independant of path taken
    for any process ΔU = nCvΔT.

    Now let us consider a isobaric process
    In this case Q = nCpΔT.

    Now taking the definition of work to be W = pΔV
    and using the defition of the equation of state pΔV = nRΔT

    then W = nRΔT

    Now consider the first law of thermodynamics again

    ΔU= Q - W

    then nCvΔT = nCpΔT + nRΔT

    leaving us with Cv = Cp + R
     
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